Biophysics: Problems II
[2.1] The Reynold’s number Re characterizes the relative importance of viscosity compared to
inertial effects in an incompressible Newtonian fluid such as water. In the case of an object
having characteristic length scale L moving in a fluid with characteristic speed U ,
Re =
U Lρ
η
where ρ is the density of the fluid and η is its viscosity. If Re 1 then viscosity dominates.
(a) Estimate Re for a bacterium of length L ∼ 1µm moving at a speed U ∼ 30µms−1 in
water at room tempertaure, for which η ∼ 10−3 kgm−1 s−1 and ρ ∼ 103 kgm−2 .
(b) Repeat the calculation for a whale with L ∼ 30m and U ∼ 10ms−1 .
[2.2] Poiseuille flow through a cylinder. In the low Reynold’s number regime, the flow of an
incompressible Newtonian liquid can be described by Stoke’s equation
ρ
∂v
= ρf − ∇p + η∇2 v
∂t
where v(r, t) is the velocity of the fluid at position r, f is an external force such as gravity
and p is the pressure.
(a) Find the steady–state solution to Stoke’s equation for flow through a circular cylinder of
radius R and length L. Take the axis of cylinder to be along the z–axis and use cylindrical
polar coordinates with the boundary condition vz (z, R) = 0. Assume that there is a uniform
pressure drop in the z–direction and neglect any external forces f .
(b) Calculate the total volume flow rate Q of fluid through the cylinder.
(c) Use the result of part (b) to estimate the corresponding flow rate per unit area of a
membrane, which consists of n parallel capillaries per unit area. Each capillary can be treated
as a cylinder of diameter d and length L.
[2.3] Bacterial swimming. Rotational diffusion affects the precision of a bacterium swimming
in a straight line. Let p(θ, t) be the probability that the bacterium is pointing in the direction
θ at time t. Suppose that p evolves according to the rotational diffusion equation
∂p(θ, t)
∂ 2 p(θ, t)
,
=D
∂t
∂θ2
with initial condition p(θ, 0) = δ(θ).
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p(2π, t) = p(0, t)
(a) Solve the rotational diffusion equation and hence show that for sufficiently small t,
θ2 ≈ 2Dt
(b) The diffusion constant satisfies an Einstein relation Dξ = kB T where ξ is the rotational
drag coefficient. In the case of a rotating sphere of radius R, it can be shown that ξ = 8πηR3 .
Estimate D for a bacterium of radius R moving in water at room temperature for which
η ∼ 10−3 kgm−1 s−1 and kB T ∼ 4.1 × 10−21 J. How long will it take the bacterium to wander
off its original direction with a standard deviation of 30o ?
(c) Suppose that the bacterium is propelling itself with speed u = 1µms−1 . At time t = 0, it
suddenly stops swimming and coasts to a stop due to the Stoke’s drag force according to the
equation
du
= −ξu, ξ = 6πηR
dt
How far does it travel before it comes to a complete stop?
m
[2.4] Dynamic strength of adhesion molecules. Biological adhesion is often dominated by
specific receptor/ligand molecules, whose strength depends on the rate at which these bonds
are loaded by a force. On laboratory time scales, Brownian-thermal excitations fully explore
the chemical energy landscape of a bond, and barriers along optimal pathways for dissociation
determine bond lifetime. Under external force, barriers in the energy landscape are lowered
and bond lifetime shortens. When isolated bonds are ruptured under steady ramps of force,
barriers diminish in time and, thus, rupture force depends on rate of loading (= force/time).
Measured over an enormous range of loading rates, the most frequent forces for failure plotted
on a scale of log (loading rate) establish a dynamic spectrum of bond strength that images the
prominent energy barriers traversed along the force-driven pathway. In this way, the so–called
method of dynamic force spectroscopy probes the inner world of molecular interactions.
Suppose that the rupture of a bond occurs via thermally assisted crossing of an activation barrier, which is diminished as the applied force increases. Assuming a one–dimensional
description for the reaction coordinate x, the rupture rate is taken to have the form
kof f (t) = k0 ef (t)∆x/kB T
where ∆x = xb − x0 is the distance from the minimum of the reaction energy E(x) to the
activation barrier at xb . In the absence of a force the spontaneous rate is k0 . Let S(t) be the
probability that the bond is still attached at time t with
dS
= −kof f (t)S(t)
dt
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koff(t)
xb
Eb - f(t)xb
-f(t)x
Then p(t) = −dS/dt is the probability density for rupture times between t → t + dt.
(a) For linear loading, f (t) = r0 t, where r0 is a constant loading rate, show that
p(y) = exp (y − [exp(y) − 1]/r0 ) /r̂0
with y = r0 t/fb , r̂0 = r0 /k0 fb and fb = kB T /∆x.
(b) Show that for r̂0 > 1 the distribution has a peak at y ∗ = log(r̂0 ). This establishes that
the most frequent bond lifetime depends logarithmically on loading rate.
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